y = (sin 2x + cot x + 2)2 at x = π/2
Given:
y = (sin2x + cotx + 2)2at x =
First, we have to find of given function, f(x),i.e, to find the derivative of f(x)
(xn) = n.xn – 1
The Slope of the tangent is
⇒ y = (sin2x + cotx + 2)2
= 2(sin2x + cotx + 2)2 – 1(sin2x) + (cotx) + (2)}
= 2(sin2x + cotx + 2)(cos2x) + ( – cosec2x) + (0)}
(sinx) = cosx
(cotx) = – cosec2x
⇒ = 2(sin2x + cotx + 2)(2cos2x – cosec2x)
Since, x =
= 2 (sin2() + cot() + 2)(2cos2() – cosec2())
= 2 (sin() + cot() + 2) (2cos() – cosec2())
= 2 (0 + 0 + 2)(2( – 1) – 1)
sin() = 0, cos() = – 1
cot() = 0,cosec() = 1
⇒ = 2(2)( – 2 – 1)
⇒ = 4 – 3
⇒ = – 12
The Slope of the tangent at x = is – 12
⇒ The Slope of the normal =
⇒ The Slope of the normal =
⇒ The Slope of the normal =
⇒ The Slope of the normal =